AP and GP crossover

Problem Statement

A sequence is both arithmetic and geometric. Given $a_1=6$ and $a_3=6$ , find all possible sequences.

Hints

Let AP difference be $d$ and GP ratio be $r$ . Use both models on $a_3$ .

Solutions

AP gives

a_3=a_1+2d=6+2d=6 \implies d=0,

so AP is constant: $a_n=6$ . For GP,

a_3=a_1r^2=6r^2=6 \implies r^2=1 \implies r=\pm 1.

If $r=-1$ , terms alternate $6,-6,6,\ldots$ and are not arithmetic. Hence the only sequence that is both AP and GP is

a_n=6\quad \text{for all }n,

with $d=0$ and $r=1$ .

Further Readings

Connect with me